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KengaRu [80]
3 years ago
8

Does a ball ever bounce back to it's drop height?

Physics
2 answers:
Valentin [98]3 years ago
5 0

Answer: No

Explanation: Once you bounce the ball it will go higher and then it will slowly come back down

ELEN [110]3 years ago
3 0

Answer:

depends the material

Explanation:

if its rubber yes for sure

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a coil has n turns enclosing an area of a. in a physics laboratory experiment, the coil is rotated during the time interval δt f
Mrac [35]
The correct answer for the question that is being presented above is this one:

Phi = BAsin(theta) 
<span>1. Phi(i) = BA </span>
<span>2. Phi(f) = 0 </span>
3. EMF = N(phi(i)-phi(f))/deltat

Here are the follow-up questions:

<span>1. What is the total magnitude Phi_initial of the magnetic flux through the coil before it is rotated? </span>
<span>2. What is the magnitude Phi_final of the total magnetic flux through the coil after it is rotated? </span>
<span>3. What is the magnitude of the average emf induced in the coil?</span>
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Sound waves travel faster through air than through solids.<br> a. True<br> b. False
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I believe the answer is true
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A box measures 20cm by 10cm by 5cm
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length × width × height

20 cm × 10 cm × 5 cm = 1,000 cm

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3 years ago
Read 2 more answers
Crushing, stirring, heating are all methods to increase the rate of ______________ a solute in a solvent.
Fofino [41]

Answer: mixing

Explanation:

4 0
3 years ago
Consider two soap bubbles with radius r1 and r2 (r1 &lt;r2) connected via a valve. What happens if we open the valve​
Sedaia [141]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

3 0
3 years ago
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